A point object is kept between a plane mirror and a concave mirror facing each other. The distance between the mirrors is 22.5 cm . The distance between the mirrors is 20 cm. what should be the distance of the object from the concave mirror so that after two successive reflections the final image is formed on the object itself ? (consider first reflection from concave mirror).

Options

A. 5 cm
B. 15 cm
C. 10 cm
D. 7.5 cm

Answer

To solve the problem step by step, we need to find the distance of the object from the concave mirror such that after two successive reflections, the final image coincides with the object itself.

### Step 1: Understand the Setup
We have a concave mirror and a plane mirror facing each other, with a distance of 22.5 cm between them. We denote the distance of the object from the concave mirror as \( x \).

### Step 2: Determine the Position of the Object and Mirrors
Let’s denote:
- Distance between the concave mirror and the plane mirror = 22.5 cm
- Distance of the object from the concave mirror = \( x \)
- Therefore, the distance of the object from the plane mirror = \( 22.5 - x \)

### Step 3: First Reflection from the Concave Mirror
Using the mirror formula for the concave mirror:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) is the focal length of the concave mirror. Given that the distance between the mirrors is 20 cm, we can assume the focal length \( f = -10 \) cm (since it's a concave mirror).
- \( u = -x \) (object distance is negative in mirror convention)
- \( v \) is the image distance from the concave mirror.

Substituting into the mirror formula:
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-x} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{x} \]
\[ \frac{1}{v} = \frac{x - 10}{10x} \]
Thus,
\[ v = \frac{10x}{x - 10} \]

### Step 5: Image Formation by the Plane Mirror
The image formed by the concave mirror (let's call it \( I_1 \)) acts as a virtual object for the plane mirror. The distance of \( I_1 \) from the plane mirror is:
\[ \text{Distance from plane mirror} = 22.5 - v \]

The image formed by the plane mirror (let's call it \( I_2 \)) will be at the same distance behind the plane mirror:
\[ \text{Distance of } I_2 = 22.5 - v \]

### Step 6: Second Reflection from the Concave Mirror
The distance of \( I_2 \) from the concave mirror is:
\[ \text{Distance from concave mirror} = 22.5 - v - 22.5 = -v \]
Now, we need to find the distance of the object from the concave mirror such that \( I_2 \) coincides with the object \( O \). Thus, we have:
\[ x = 22.5 - v \]

### Step 7: Substitute for \( v \)
Substituting the expression for \( v \):
\[ x = 22.5 - \frac{10x}{x - 10} \]

### Step 8: Solve for \( x \)
Cross-multiplying gives:
\[ x(x - 10) = 22.5(x - 10) - 10x \]
Expanding and simplifying:
\[ x^2 - 10x = 22.5x - 225 - 10x \]
\[ x^2 - 22.5x + 225 = 0 \]

### Step 9: Factor the Quadratic Equation
The quadratic equation can be factored or solved using the quadratic formula:
\[ x^2 - 15x - 30x + 450 = 0 \]
Factoring gives:
\[ (x - 15)(x - 30) = 0 \]
Thus, \( x = 15 \) or \( x = 30 \).

### Step 10: Determine Valid Solution
Since the distance between the mirrors is 22.5 cm, \( x = 30 \) cm is not possible. Therefore, the valid solution is:
\[ x = 15 \text{ cm} \]

### Final Answer
The distance of the object from the concave mirror should be **15 cm**.